Question

A solid copper sphere (density $\rho$ and specific heat capacity $c$) of radius $r$ at an initial temperature $200 \,K$ is suspended inside a chamber whose walls are at almost $0\, K$. The time required (in $\mu s$ ) for the temperature of the sphere to drop to $100\, K$ is

• A. $\frac{72}{7} \frac{r \rho c}{\sigma}$
• B. $\frac{7}{72} \frac{r \rho c}{\sigma}$
• C. $\frac{27}{7} \frac{r \rho c}{\sigma}$
• D. $\frac{7}{27} \frac{r \rho c}{\sigma}$

Answer 1

Answer: (B)

$\frac{d T}{d t}=\frac{\sigma A}{m c J}\left(T^{4}-T_{0}^{4}\right)$ [In the given problem, fall in temperature of body $d T=(200-100)=100\, K$, temp. of surrounding $T_{0}=0\, K$, Initial temperature of body $T=200 \,K$ ]
$\frac{100}{d t} =\frac{\sigma 4 \pi r^{2}}{\frac{4}{3} \pi r^{3} \rho c J}\left(200^{4}-0^{4}\right)$
$\Rightarrow d t =\frac{r \rho c J}{48 \sigma} \times 10^{-6} \cdot s =\frac{r \rho c}{\sigma} \cdot \frac{4.2}{48} \times 10^{-6} $
$=\frac{7}{80} \frac{r \rho c}{\sigma} \mu s \simeq \frac{7}{72} \frac{r \rho c}{\sigma} \mu s $[ As $ J =4.2]$